Question

The sum of the terms of an infinite geometric sequence with common ratio r(0<r<1) is 15 and the sum of its squares is 45. Find the sequence.
Full Handwritten answers only​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$1.Let \: 'a' \: is \: first \:term \: of \: a \: G.P$

$Common \: ratio = r \: ( given )$

$\blue { Sum \: of \: infinite \: sequence \: S_{\infty}}$

$\blue{ \frac{a}{1-r}} = 15\: --(1)$

$\implies a = 15(1-r) \: ---(2)$

$2. If \: doing \: squares \: of \: each \:term$

$\:then\: the \: new \: G.P \: is$

$a^{2},a^{2}r^{2}, (ar^{2}), \ldots$

$\pink { Sum \: of \: infinite \: sequence }$

$\frac{a^{2}}{1-r^{2}} = 45 \: ( given )$

$\implies \frac{a}{1-r} \times \frac{a}{1+r} = 45$

$\implies 15 \times \frac{a}{1+r} = 45 \: [ From \:(1) ]$

$\implies \frac{a}{1+r} = \frac{45}{15}$

$\implies \frac{a}{1+r} = 3$

$\implies a = 3(1+r) \: ---(3)$

/* From equation (2) and (3), we get */

$15(1-r) = 3(1+r)$

$\implies 5(1-r) = 1+ r$

$\implies 5 - 5r = 1 + r$

$\implies - 5r - r = 1 - 5$

$\implies - 6r = - 4$

$\implies r = \frac{- 4}{-6}$

$\implies r = \frac{2}{3} \:---(4)$

$\implies put \: r = \frac{2}{3} \: equation \:(2)$

$a = 15 \big(1 - \frac{2}{3} \big)$

$\implies a = 15 \big( \frac{3-2}{3}\big)$

$\implies a = 15 \times \frac{1}{3}$

$\implies a = 5$

Therefore.,

$\blue { a = 5 \: and \: r = \frac{2}{3} }$

$\red{ Required \: G.P : }$

$\green{ 5, \frac{10}{3} , \frac{20}{9}\ldots }$

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